If the probability distribution function for the random variable X is f(x), then
first calculate E(X) = integral of x*f(x)dx over the whole real line.
Noxt calculate E(X2) = integral of x2*f(x)dx over the whole real line.
Then Variance(X) = E(X2) - [E(X)]2
and finally, SD(X) = sqrt[Variance(X)].
The mean and standard deviation do not, by themselves, provide enough information to calculate probability. You also need to know the distribution of the variable in question.
The Poisson distribution is a discrete distribution, with random variable k, related to the number events. The discrete probability function (probability mass function) is given as: f(k; L) where L (lambda) is the mean and square root of lambda is the standard deviation, as given in the link below: http://en.wikipedia.org/wiki/Poisson_distribution
It depends what you're asking. The question is extremely unclear. Accuracy of what exactly? Even in the realm of statistics an entire book could be written to address such an ambiguous question (to answer a myriad of possible questions). If you simply are asking what the relationship between the probability that something will occur given the know distribution of outcomes (such as a normal distribution), the mean of that that distribution, and the the standard deviation, then the standard deviation as a represents the spread of the curve of probability. This means that if you had a cure where 0 was the mean, and 3 was the standard deviation, the likelihood of observing a value of 12 (or -12) would be likely inaccurate if that was your prediction. However, if you had a mean of 0 and a standard deviation of 100, the likelihood of observing of a 12 (or -12) would be quite likely. This is simply because the standard deviation provides a simple representation of the horizontal spread of probability on the x-axis.
For data sets having a normal distribution, the following properties depend on the mean and the standard deviation. This is known as the Empirical rule. About 68% of all values fall within 1 standard deviation of the mean About 95% of all values fall within 2 standard deviation of the mean About 99.7% of all values fall within 3 standard deviation of the mean. So given any value and given the mean and standard deviation, one can say right away where that value is compared to 60, 95 and 99 percent of the other values. The mean of the any distribution is a measure of centrality, but in case of the normal distribution, it is equal to the mode and median of the distribtion. The standard deviation is a measure of data dispersion or variability. In the case of the normal distribution, the mean and the standard deviation are the two parameters of the distribution, therefore they completely define the distribution. See: http://en.wikipedia.org/wiki/Normal_distribution
idk about normal distribution but for Mean "M" = (overall sum of "x") / "n" frequency distribution: 'M' = Overall sum of (' x ' * ' f ') / overall sum of ( ' f ' ) M = Mean x = Mid Point f = frequiency n = number of variables ALL FOR STANDARD DEVIATION * * * * * A general Normal distribution is usually described in terms of its parameters, and given as N(mu, sigma2) where mu is the mean and sigma is the standard deviation. The STANDARD Normal distribution is the N(0, 1) distribution, that is, it has mean = 0 and variance (or standard deviation) = 1.
Square the standard deviation and you will have the variance.
In general you cannot. You will need to know more about the distribution of the variable - you cannot assume that the distribution is uniform or Normal.
From the related link, you can read directly the probability that Z is less than 1.51 is 0.9345.
In order to answer this question, you will need to provide the standard deviation.
It is a variable that can take a number of different values. The probability that it takes a value in any given range is determined by a random process and the value of that probability is given by the probability distribution function.It is a variable that can take a number of different values. The probability that it takes a value in any given range is determined by a random process and the value of that probability is given by the probability distribution function.It is a variable that can take a number of different values. The probability that it takes a value in any given range is determined by a random process and the value of that probability is given by the probability distribution function.It is a variable that can take a number of different values. The probability that it takes a value in any given range is determined by a random process and the value of that probability is given by the probability distribution function.
If G is a random variable representing the gestation period, an it is assumed to have a normal distribution with the given mean and standard deviation, then Prob(250 < G < 282) = Prob(-1 < Z < 1) where Z has the standard normal distribution. = 0.6827 = 68.3%
A single number, such as 478912, always has a standard deviation of 0.
Sampling distribution is the probability distribution of a given sample statistic. For example, the sample mean. We could take many samples of size k and look at the mean of each of those. The means would form a distribution and that distribution has a mean, a variance and standard deviation. Now the population only has one mean, so we can't do this. Population distribution can refer to how some quality of the population is distributed among the population.
The Empirical Rule states that 68% of the data falls within 1 standard deviation from the mean. Since 1000 data values are given, take .68*1000 and you have 680 values are within 1 standard deviation from the mean.
Standard deviations are measures of data distributions. Therefore, a single number cannot have meaningful standard deviation.
Assuming the returns are nomally distributed, the probability is 0.1575.
standard deviation is the square roots of variance, a measure of spread or variability of data . it is given by (variance)^1/2
A test using relative errors comparing a frequency table to the expected counts determined using a given probability distribution; the null hypothesis is that the given probability distribution fits the data's distribution.
All Gaussian (Normal) distributions are determined by just two parameters: their mean and standard deviation. Thus, given any variable, X, that has a Gaussian distribution with mean (m) and standard deviation (s), the Z transform, which is Z = (X - m)/s has a standard normal [or N(0,1)] distribution. This is tabulated and the tables can be used for testing statistical hypothesis about Normally distributed variables.
A standard deviation in statistics is the amount at which a large number of given values in a set might deviate from the average. A percentile deviation represents this deviation as a percentage of the range.
A mathematical definition of a standard normal distribution is given in the related link. A standard normal distribution is a normal distribution with a mean of 0 and a variance of 1.
A standard deviation calculator allows the user to find the mean spread away from the mean in a statistical environment. Most users needing to find the standard deviation are in the statistics field. Usually, the data set will be given and must be typed into the calculator. The standard deviation calculator will then give the standard deviation of the data. In order to find the variance of the data, simply square the answer.
49.30179172 is the standard deviation and 52 is the mean.
You do not solve a standard normal distribution. It is not a question nor an equation or inequality to be solved. You can answer questions using the standard normal distribution but what you do depends on the question and on what information is given.
This depends on what information you have. If you know the success probability and the total number of observations, you can use the given formulas. Most of the time, this is the case. If you have data or experience which allow you to estimate the parameters, it may sometimes happen that you work like this. This mostly happens when n is very large and p very small which results in an approximation with the Poisson distribution.